Spread the love

Commercial Unit of Electrical Energy | (BOTU)

Commercial Unit of Electrical Energy :- The cost of electricity consumed in the factories, houses and for many other works is calculated on the basis of a unit called kilowatt hour(kWh).

What is Commercial Unit of Electrical Energy | (BOTU) | kWh

Commercial Unit of Electrical Energy i.e., “Board of trade unit” (BOTU) or kWh is actually a unit of electrical energy.

1 kWh or 1 B.O.T.U or 1 Unit :- “It is the amount of electrical energy which is consumed in one hour in an electric circuit of power 1 KW”.

1 Kilowatt-hour = 1 Kilowatt × 1 hour = 1000 watt × 3600 second

1 Kilowatt-hour = 3.6 × 10⁶ J

If I current flows in an electric circuit for t hours at a voltage V, then electrical energy consumed in the circuit,

E = Power(P) × time(t)

E = Pt = (VI × t) watt-hour

$\displaystyle E=\frac{VI}{1000}\times t\text{ Kilowatt}\times \text{hour}$

$\displaystyle E=\frac{VIt}{1000}\text{Kilowatt}\times \text{hour(unit)}$

$\displaystyle Therefore\text{ }number\text{ }of\text{ }units\text{ }consumed\text{ }=~\frac{volt\times ampere\times hour}{1000}$

$\displaystyle Number\text{ }of\text{ }units\text{ }consumed\text{ }=~\left( \frac{Power(\text{in watt})\times time(in\text{ hour})}{1000} \right)$

Power Rating of appliances

Each electrical appliance or instrument (for example bulb) is marked with watt(W) and volt(V) under which the instrument can be used properly. For example, if 100 W, 220 V is marked on a bulb then it means that on applying a potential difference of 220 volt across the ends of the bulb the electric power 100W is consumed (or energy consumption is 100 joule per second). 220 V mark on the bulb means that maximum potential difference that can be applied is 220 V. On applying a potential difference more than 220 V, the bulb cannot remain safe, i.e., it may fuse.

To find resistance and maximum safe current for appliance

Let the rating of a bulb is : Power – P(in watt) , Potential difference – V(in volts)

Resistance of the filament of the bulb, $\displaystyle R=\frac{{{V}^{2}}}{P}$ & …..(1)

Maximum safe current, $\displaystyle I=\frac{P}{V}$ …..(2)

From equation (1), it is clear that :

“For two electrical appliance of same operating voltage(potential difference) the resistance of high power(wattage) electrical appliance is less and the resistance of low power(wattage) electrical appliance is more”.

To understand this let us consider two bulbs as shown in figure below :-

Resistance of Bulb 1, $\displaystyle {{R}_{1}}=\frac{{{V}^{2}}}{P}=\frac{{{220}^{2}}}{100}=484\Omega$

Resistance of Bulb 2, $\displaystyle {{R}_{2}}=\frac{{{V}^{2}}}{P}=\frac{{{220}^{2}}}{50}=968\Omega$

⇒ Resistance of 50 W bulb > Resistance of 100 W bulb

Illumination of bulb

The brightness of a bulb depends on the temperature of it’s filament. The temperature depends on the heat produced. That in turn depends on the power being dissipated in the filament.

Power is given by P = V × I. Therefore an increase in either one will increase the brightness.

In practice, the filament resistance is also a function of temperature, it increases somewhat as the temperature increases.

But on the whole, we control voltage and when increased, it causes the current to go up according to Ohm’s law and the brightness increases.

So they are interrelated, brightness goes up because of both.

If the temperature of the filament reaches its melting point due to the heat produced in the bulb, then the filament of the bulb breaks down and flow of current stops. This position is called fusing of the bulb.

Parallel combination of bulbs

If many bulbs of different power i.e., 25W, 60W, 100W etc, are connected in parallel, then the potential difference across each bulb remains same. As $\displaystyle P=\frac{{{V}^{2}}}{R}$ , maximum power is consumed in the bulb of minimum resistance and minimum power is consumed in the bulb of maximum resistance.

Therefore the bulb of minimum resistance illuminates with maximum brightness and the bulb of maximum resistance illuminates with minimum brightness.

So a bulb of 100W will give more brightness than a bulb of 50W.

Total electric power consumed in parallel combination is given by :-

P = VI = V(V/R_eq) = V²/R_eq

$\displaystyle P={{V}^{2}}\left[ \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+..... \right]$

$\displaystyle P=\frac{{{V}^{2}}}{{{R}_{1}}}+\frac{{{V}^{2}}}{{{R}_{2}}}+\frac{{{V}^{2}}}{{{R}_{3}}}+.....$

⇒ $\displaystyle P={{P}_{1}}+{{P}_{2}}+{{P}_{3}}+.....$

Series combination of bulbs

If many bulbs of different power i.e., 25W, 60W, 100W etc, are connected in series, then electric current(I) passing through each bulb remains same. As $\displaystyle P={{I}^{2}}R$ , maximum power is consumed in the bulb of maximum resistance and minimum power is consumed in the bulb of minimum resistance.

Therefore the bulb of minimum resistance illuminates with minimum brightness and the bulb of maximum resistance illuminates with maximum brightness.

So a bulb of 50W will give more brightness than a bulb of 100W.

Total electric power consumed in series combination is given by :-

P = VI = V(V/R_eq) = V²/R_eq

$\displaystyle P=\frac{{{V}^{2}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{2}}+.....}$

$\displaystyle \Rightarrow \frac{1}{P}=\frac{{{R}_{1}}+{{R}_{2}}+{{R}_{2}}+.....}{{{V}^{2}}}$

$\displaystyle \frac{1}{P}=\frac{{{R}_{1}}}{{{V}^{2}}}+\frac{{{R}_{2}}}{{{V}^{2}}}+\frac{{{R}_{3}}}{{{V}^{2}}}.....$

$\displaystyle \frac{1}{P}=\frac{1}{{}^{{{V}^{2}}}\!\!\diagup\!\!{}_{{{R}_{1}}}\;}+\frac{1}{{}^{{{V}^{2}}}\!\!\diagup\!\!{}_{{{R}_{1}}}\;}+\frac{1}{{}^{{{V}^{2}}}\!\!\diagup\!\!{}_{{{R}_{1}}}\;}+.....$

⇒ $\displaystyle \frac{1}{P}=\frac{1}{{{P}_{1}}}+\frac{1}{{{P}_{2}}}+\frac{1}{{{P}_{3}}}+.....$

Transmission of power to distant places

When electric power from power station to houses and factories is transmitted through the cables of resistance R then the electrical power dissipated as heat is given by I²R.

Now if P power is to be transmitted at V volt then,

$\displaystyle P=VI$

$\displaystyle I=\frac{P}{V}$

Therefore the electrical power dissipated as heat will be $\displaystyle \frac{{{P}^{2}}}{{{V}^{2}}}\times R$

Now the power to be transmitted P and the resistance of the cables R is constant, therefore

$\displaystyle Electrical\text{ }power\text{ }dissipated\text{ }as\text{ }heat\propto \frac{1}{{{V}^{2}}}$

So, when the power is transmitted at high voltage then power loss will be less. This is the reason why transmission of power is done at high voltages.

Commercial Unit of Electrical Energy | (BOTU)

Commercial Unit of Electrical Energy | (BOTU)

What is Commercial Unit of Electrical Energy | (BOTU) | kWh

Like this:

Leave a Reply Cancel reply